- #1

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help please

- Thread starter sundrops
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- #1

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help please

- #2

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- #3

cepheid

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The x-component of the v_0 is unaffected throughout the journey. Therefore, you don't have a clue yet what to do with it. It is constant, so you need to know the time required for the ball to get there. That depends on how high it needs to go before stopping (the vertical component). So, just note for now that the x component is given by :

[tex] v_0\cos(38^o) [/tex]

and ignore it for the time being.

Now, the vertical component of the velocity:

[tex] v_0\sin(38^o) [/tex]

is affected by the gravitational force, which causes a constant downward acceleration, which slows the ball's velocity by 9.81 metres per second, every second. So, how do you solve for the required initial vertical velocity, and the time needed to get the ball up to that height? There are three equations...one gives the vertical component of the velocity as a function of time, (which is useful to you, because you know the final vertical velocity is zero), the other gives the height h(t) as a function of time, in terms of the vertical velocity and and accelaration. The third is the fact that:

[tex] [v_0\cos(38^o)]t [/tex] must be equal to three metres.

Hopefully you know which kinematic formulas I'm talking about.

- #4

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Vxo = Vx(cos38) Vyo=Vy(sin38)

combining the two I found,

y=(tan38)x - (g / 2(Vo^2)(cos38)^2)x^2

however my answer was wrong.

( I tried your method aswell cepheid - but it was long and complicated - I think I got lost in the numbers along the way.) :s

- #5

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did you mean

[tex]y = tan(38)x - \frac{g x^2}{2 v_{0}^2 cos^2(28)) [/tex]

if yes, your answer should be fine, what numerical v you got?

- #6

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and whats frac?

I found Vo = 6.115 m/s and it was wrong...:(

- #7

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that was a typo

here you go

[tex]y = tan(38)x - \frac{g x^2}{2 v_{0}^2 cos^2(38)} [/tex]

let me check... 1 minute

here you go

[tex]y = tan(38)x - \frac{g x^2}{2 v_{0}^2 cos^2(38)} [/tex]

let me check... 1 minute

Last edited:

- #8

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how did you write it out so clearly??

- #9

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i don't get your answer... let me try to figure out what you diid wrong...

- #10

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was it just a calculation error then? this is the right equation to use to solve the problem?

- #11

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I'll try it again - maybe i din't put brackets around certain #'s in my calculator or something...

- #12

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basically I plugged in:

x = 7.0m

y = 0.3m

g = 9.8m/s^2

and solved for Vo

x = 7.0m

y = 0.3m

g = 9.8m/s^2

and solved for Vo

- #13

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what is your Vo equal to... you might either have some algebra error or calculatioin error

- #14

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31.95m/s ??

- #15

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does that sound more feasible?

- #16

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i got 5 point something...

sory, can't give you the answer, try again

sory, can't give you the answer, try again

- #17

- 55

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that answer didn't work either

what am I doing wrong?

~ I only have 2 tries left :S

what am I doing wrong?

~ I only have 2 tries left :S

- #18

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ok

well at least I know the approximate range of the answer right? hehe

well at least I know the approximate range of the answer right? hehe

- #19

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Vo = \sqrt{(tan(38)(7.0m) - \frac{(-9.8m/s^2) (7.0m)^2}{2 (0.3m) cos^2(38)}}

- #20

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here's how I manipulated the equation:

Vo = Square root of { tan(38)x - ((-g)x^2 / 2ycos(38)^2) }

- #21

- 609

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completely wrong. i believe you can do this problem your own.... just simple algebra.

- #22

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- #23

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g wpuld be a negative number right?

(-9.8m/s^2 as opposed to 9.8m/s^2)

(-9.8m/s^2 as opposed to 9.8m/s^2)

- #24

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the negative sign has already been taken care of in the equation, g is positive

- #25

- 55

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okay - that makes sense.

I keep getting 8.65m/s

I keep getting 8.65m/s

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